# The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point A.

Let the height of the tower is = h (m) Distance of point B from foot of the tower is = (m)

tan 30° =  = √3 h = 20+ -----------(1)

In ∆DCB,

tan 60° = √3 = h = √3 ----------(2)

On substituting value of h from eqn. (2) in eqn. (1)

√3× √3 = 20 + 3 20+ 3 - = 20 = 10

Therefore distance of point A from tower is

AC = AB+BC

AC = 20+10 30

Ac = 30 m.

Now substituting value of in eqn. (1) h = 20+10 30 17.32 m.

Therefore height of tower is 17.32 m.

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